How to find out the angle here Let $ P $ be an interior point of $ ∆ABC $ , such that $ Q $ and $ R  $ are the reflections of $ P $ in $ AB $ and $ AC $ respectively.
Also it is given that $ Q,A,R $ are collinear.
Then i need to find out $ \angle A $ . How to do it?
(I'm also having trouble to draw the figure)
 A: Why are you having problems drawing this? Are you starting with the triangle, and then can't get the collinearity in place?
Consider this: the question is asking you to determine an angle, but giving you pretty little information apart from that. This suggests that starting with an arbitrary angle and trying to create the drawing from that is likely to fail, because the construction only works for some very specific angle which you have yet to find.
There are two approaches I'd suggest. One is trying out “special” angles. Things like $90°, 60°, 45°, 120°, 30°$. Chances are one of them will make the pieces fall into place. The other approach would be to more closely examine how the construction fails depending on $\angle A$. Does it fail “more strongly” for some angles than for others? In that case go in the direction that looks closer to a solution. Does it fail one way for some angles and the opposite way for others? Then the solution you are after is likely in between these two. That way you can come up with a pretty good guess as to what the angle would have to be.
Depending on how rigorous you have to be, your next step might be proving that the chosen angle does lead (or can lead) to the described configuration. That shouldn't be too hard. Trickier would be the task of showing that no other angle can lead to said configuration. If you need that done rigorously, I guess I'd try for an algebraic description of the problem, since there it's easier to make sure you haven't missed some corner case.
I've deliberately not included the angle or a drawing. I think that with these steps you should be able to find the answer. If you do, feel free to answer your own question, providing the angle, a drawing and at least an intuitive description of what's going on.
A: If $Q'$ is the intersection of $PQ$ with $AB$ and $R'$ is the point of intersection of $PR$ with $BC$
then $\angle{PAQ'} = \angle{Q'AQ} = \alpha$ and $\angle{PAR'} = \angle{R'AR} = \beta$
So $\angle{BAC} = \angle{R'AQ'} = \alpha + \beta$
But since $RAQ$ is a straight line, with the angle $2(\alpha + \beta)$
where must have that
$$\angle{BAC} = \alpha + \beta = 90^{\circ}$$
